You actually already proved the difficult part.
If $J$ is maximal, then by the correspondence of ideals you mention $J'$ has to be maximal, too. If it weren't, there would be a proper ideal $T'\supset J'$, which would give a proper ideal $T\supset J$.
Now assume $J$ prime and let $a'b'\in J'$, where $a',b'$ are the classes in $R/I$ of some $a,b\in R$. This means that there are $i_a,i_b\in I$ such that $(a+i_a)(b+i_b)=ab+ai_b+bi_a+i_ai_b\in J+I=J$, as $I\subseteq J$. Thus $ab\in J$ and, say, $a\in J$. Hence $a'\in J'$, so $J'$ is prime.
Finally, suppose $J$ radical and let $(a')^r\in J'$ for some $a\in R$, $r\in\Bbb N$. Then there are some $i_a,i\in I$ such that, by binomial expansion and since $I$ is an ideal, $(a+i_a)^r=a^r+i\in J+I=J$. Hence $a^r\in J$, so $a\in J$. Therefore $a'\in J'$ and $J'$ is radical.
Unfortunately, this is impossible! If the intersection of a pair of primary ideals was primary, then the intersection of any two prime ideals would be prime. But we know this is not the case. It's a misstep to just ignore the added condition that both ideals be $P$ primary.
In the integers, for example, $(2)$ and $(3)$ are primary (actually prime!) ideals, but their intersection $(6)$ is not a primary ideal.
You will need to make full use that they are $P$-primary. The previous example I gave is not a contradiction to the $P$-primary statement since the $P$ primary ideals in $\Bbb Z$ form a chain.
So starting over again with your notation, add in additionally that $Rad(P_1)=Rad(P_2)=P$. Notice that right away you have $a,b\in P=Rad(P_1)=Rad(P_2)$, which is critical. (Solve this and do $n$-ary intersections with induction.)
Best Answer
Consider a commutative unital ring $R$ and some prime ideals $P_1, \dots, P_n.$
Claim. We have that $I = P_1 \cap \cdots \cap P_n$ is radical, i.e., $\sqrt I = I.$
Proof. Considering that radicals distribute over intersections, we have that $$\sqrt I = \sqrt{P_1 \cap \cdots \cap P_n} = \sqrt{P_1} \cap \cdots \cap \sqrt{P_n}.$$ But prime ideals are radical, so $\sqrt{P_i} = P_i$ implies that $\sqrt I = I.$ QED.